学数答题160909-取整函数
题160908(14分)对正整数$n$及一切实数$x$,求证:
$\left[ x \right]+\left[ x+\dfrac{1}{n} \right]+\left[ x+\dfrac{2}{n} \right]+\cdots +\left[ x+\dfrac{n-1}{n} \right]=\left[ nx \right]$.
注:$\left[ x \right]$为取整函数,即不超过$x$的最大整数.
题目来源:重要恒等式——厄尔米特恒等式,特别地,$n=2$时有$\left[ x \right]+\left[ x+\dfrac{1}{2} \right]=\left[ 2x \right]$.
解:对任意的正整数$n$,构造函数$f\left( x \right)=\left[ nx \right]-\left[ x \right]-\left[ x+\dfrac{1}{n} \right]-\left[ x+\dfrac{2}{n} \right]-\cdots -\left[ x+\dfrac{n-1}{n} \right]$,
则$f\left( x+\dfrac{1}{n} \right)$
$=\left[ nx \right]+1-\left[ x+\dfrac{1}{n} \right]-\left[ x+\dfrac{2}{n} \right]-\cdots -\left[ x+\dfrac{n-1}{n} \right]-\left( \left[ x \right]+1 \right)$
$=f\left( x \right)$
所以函数$f\left( x \right)$为周期函数,其周期$T=\dfrac{1}{n}$,
因此原命题只需证明$f\left( x \right)=0$在区间内成立即可,显然成立.
法2:设$x=\left[ x \right]+\left\{ x \right\}$,且$\left\{ x \right\}\in \left[ \dfrac{k}{n},\dfrac{k+1}{n} \right)$,$k$是非负整数.
右端$=n\left[ x \right]+\left[ n\left\{ x \right\} \right]=n\left[ x \right]+k$,
左端$=n\left[ x \right]+\left[ \left\{ x \right\}+\dfrac{1}{n} \right]+\left[ \left\{ x \right\}+\dfrac{2}{n} \right]+\cdots ++\left[ \left\{ x \right\}+\dfrac{n-1}{n} \right]$,
注意到$\left\{ x \right\}\in \left[ \dfrac{k}{n},\dfrac{k+1}{n} \right)$,显然从$\left[ \left\{ x \right\}+\dfrac{n-k}{n} \right]$项起都是$1$,
即$\left[ \left\{ x \right\}+\dfrac{n-k}{n} \right]=\left[ \left\{ x \right\}+\dfrac{n-\left( k-1 \right)}{n} \right]=\cdots =\left[ \left\{ x \right\}+\dfrac{n-1}{n} \right]=1$,
共$k$个$1$(之前其他项都是$0$).
从而左端也是$n\left[ x \right]+k$,得证.